If two numbers will be randomly chosen without replacement from $\{3, 4, 5, 6\}$, what is the probability that their product will be a multiple of 9? Express your answer as a common fraction.
Solution: There are $\binom{4}{2}=6$ possible pairs of numbers that can be chosen. None of these numbers are multiples of 9, so in order for their product to be a multiple of 9, both numbers must be a multiple of 3. The only possible pair that satisfies this is 3 and 6. Thus, the probability is $\boxed{\frac{1}{6}}$